Opamp help???

[shaneshane1]

just wanted to know what Opamps are used for, i have searched google a little and found some info, but there info is to technical and long for me to fully understand.

I understand the pin out for a basic Opamp, as for working out what the output is ment to do, im stumped

Is it anything like a transistor?

Is it anything like a votage regulator?

If im using only a 5V supply can i some how increase the output to say 10V?

I know this is a real "noob" question, but if someone can explain it to me as if i were 10 years old, i will better understand it rather than making it to hard to understand, once i get the basic understandings of how it works, i can then work the rest out for my self, Thanks!!!

 

[dknguyen]

An op-amp can be used as a regulator with additional components. It's made of a bunch of transistors so it could be made to imitate a transistor (especially a transistor wired as an amplifier).

Depending on whether an op-amp is voltage-feedback or current-feedback type, all it does is output a voltage (or current) from it's output pin until the voltage at the + input pin equals the voltage at the - input pin. You can use this characteristic to make amplifiers, filters, buffers, "ideal" diodes, regulators, mathematical operations and all sorts of stuff.

To get a 10V output from an op-amp, you have to use a 10V or higher supply, since like a linear regulator, an op-amp's maximum and minimum output is always less (and ideally equal) to the supply voltage.

You need a switching DC-DC converter or charge pump or something to turn your 5V supply into 10V.

 

[shaneshane1]

An op-amp can be used as a regulator with additional components. It's made of a bunch of transistors so it could be made to imitate a transistor (especially a transistor wired as an amplifier).

Depending on whether an op-amp is voltage-feedback or current-feedback type, all it does is output a voltage (or current) from it's output pin until the voltage at the + input pin equals the voltage at the - input pin. You can use this characteristic to make amplifiers, filters, buffers, "ideal" diodes, regulators, mathematical operations and all sorts of stuff.

To get a 10V output from an op-amp, you have to use a 10V or higher supply, since like a linear regulator, an op-amp's maximum and minimum output is always less (and ideally equal) to the supply voltage.

You need a switching DC-DC converter or charge pump or something to turn your 5V supply into 10V.

cool, thanks!!!

what would be the most basic schematic/project to start out with to get the hang of using an Opamp.

When i say "basic" i mean least amount of parts used and very easy to understand, something to do with LED's would be great

 

[dknguyen]

cool, thanks!!!

what would be the most basic schematic/project to start out with to get the hang of using an Opamp.

When i say "basic" i mean least amount of parts used and very easy to understand, something to do with LED's would be great

EIther a comparator or a buffer. They take no extra parts, just different wiring. Wiki it. You can use an LED to read the output rather than just a voltage if you wish in which case you should have a current limiting resistor (like always) and use an op-amp that can output 20mA or so.

 

[shaneshane1]

i sort of understand how it works, but not sure if im reading it correctly, if my input on the inverting input is say 1V and i have the non-inverting input tied to ground, will the output be 4V, providing that my power supply is 5V, or is that all wrong,

 

[dknguyen]

i sort of understand how it works, but not sure if im reading it correctly, if my input on the inverting input is say 1V and i have the non-inverting input tied to ground, will the output be 4V, or is that all wrong,

In this case, there is no feedback so it will act as comparator (an amplifier with pretty much infinite gain) so it will saturate at it's supply voltage limit. So if inverting = 1V, then 1V is the reference it is comparing or amplifying relative to. You tied non-inverting to ground. So as a comparator it will output a LO (as close to it's lower supply rail as possible). If you think of it as an amplifier then it will amplify the 0V (ground) relative to 1V as much as possible until it saturates:

[0V - 1V (the bias point)] x infinite gain = -infinite volts. But since it can't go that low it saturates at the lower supply rail.

 

[shaneshane1]

In this case, there is no feedback so it will act as comparator (an amplifier with pretty much infinite gain) so it will saturate at it's supply voltage limit. So if inverting = 1V, then 1V is the reference it is comparing or amplifying relative to. You tied non-inverting to ground. So as a comparator it will output a LO (as close to it's lower supply rail as possible). If you think of it as an amplifier then it will amplify the 0V (ground) relative to 1V as much as possible until it saturates:

[0V - 1V (the bias point)] x infinite gain = -infinite volts. But since it can't go that low it saturates at the lower supply rail.

Does changing the values of R1,R2 and R3 change the output voltage, or does it change the Amp output, or does it change both?


If i DONT apply a voltage to the V+(inverter input) what will the output show?

and if i DO apply 1V to the V+(inverter input) what will the output show?

 

[dknguyen]

THe amp voltage and output voltage are the same voltage. What do you mean?

CHanging the R1 and R2 will change the gain- that should answer your question about what happens if you apply 1V to the inverting input. R3 doesn't do anything except to load the signal source on the inverting input. If you don't apply a voltage the output will float and will be unpredictable and change randomly.

 

[shaneshane1]

THe amp voltage and output voltage are the same voltage. What do you mean?

CHanging the R1 and R2 will change the gain- that should answer your question about what happens if you apply 1V to the inverting input. R3 doesn't do anything except to load the signal source on the inverting input. If you don't apply a voltage the output will float and will be unpredictable and change randomly.

what i meant was is the output current(Amps) effected by the resistors?


So if i have R1 and R2 as 10K and R3 as say 4.7K, what will the output be if the input(inverter) is 1V?

 

[dknguyen]

what i meant was is the output current(Amps) effected by the resistors?


So if i have R1 and R2 as 10K and R3 as say 4.7K, what will the output be if the input(inverter) is 1V?

Input resistors do not affect the output current as long as the resistor values are in reason. If they are too low, obviously too much current might go through them for the op-amp (or even the input sources) to drive properly (create enough current to produce the appropriate voltage across them).

THe gain is -(1+R2/R1)

 

[shaneshane1]

THe gain is -(1+R2/R1)

what does the - mean, and does (1+R2/R3) = 1.1?

If so 1.1 what? Volts

 

[dknguyen]

1+R2/R3 = 2 if R2=R3. THis means the gain of the amplifier is 2. So 1V in is 2V out.

Since it has a negative sign in front (just due to the way current is flowing in the circuit and how that affects the polarity voltages being formed across the resistors), then 1V in is -2V out. BUt since the supply doesn't go as low as -2V, it just saturates at 0V.

Please learn the principles behind op-amp math and remember that all an op-amp does is drive enough output voltage (or current) so that the voltage at the inverting and non-inverting terminals are equal. As a quick intro to this concept please follow this example:
**broken link removed**
Assume the op-amp has it's power supply infinite positive voltage and infinite negative voltage so that it never saturates. You apply a voltage at Vin. Since the op-amp terminals are ideally infinite impedance, no current flows into the opamp input so no voltage drop is formed across the resistor so the voltage at Vin equals the voltage at the non-inverting input. Got it?

Now the op-amp will try and output enough voltage to make the voltages at the two input terminals equal. If the voltage at the non-inverting input is more positive than the one at the inverting input, the op-amp will SOURCE current (ie. output a positive voltage) and if the inverting is more positive than the non-inverting the op-amp will SINK current (ie. output a negative voltage). It will keep increase (or decreasing) the output voltage with this trend until the input voltages are equal. So let's assume Vin is positive (ie. above ground which is connected to the inverting terminal through resistors). THe op-amp will output a current down through the resistors to ground such that the voltage formed by this current across Rin (and therefore the inverting input voltage) is equal to the non-inverting input voltage.

BUt what's this? THis current also goes through Rf! THat means that the current flowing through the two resistors (remember no current flows into the op-amp inputs) is I=Vin/Rin to cause the voltage at the inverting input to match the voltage at the non-inverting input. BUT this same current is flowing through both resistors and the voltage drop formed across them is Vout = I*(Rin+Rf). ANd since this resistor divider is referenced to ground, then the output voltage is also referenced to ground. If we rearrange and make the two equations equal to each other, we get:

Gain = Vout/Vin = (Rin+Rf)/Rin = 1 + Rf/in

THe gain is positive. WHat is happening is the magnitude of the current is being decided by one resistor, but the output voltage is being decided by the current flowing through a LARGER resistance (the one resistor plus another resistor) so you get a voltage gain. In order for the op-amp to produce a matching voltage at the input terminals (ie. drive enough current to produce the voltage across Rin) it must have a higher voltage at it's output since this extra voltage is being dropped across the other resistor, Rf.

THat is as far as I will go. Please go through this link carefully.

Walk through this:

 

[shaneshane1]

1+R2/R3 = 2 if R2=R3. THis means the gain of the amplifier is 2. So 1V in is 2V out.

Since it has a negative sign in front (just due to the way current is flowing in the circuit and how that affects the polarity voltages being formed across the resistors), then 1V in is -2V out. BUt since the supply doesn't go as low as -2V, it just saturates at 0V.

Please learn the principles behind op-amp math and remember that all an op-amp does is drive enough output voltage (or current) so that the voltage at the inverting and non-inverting terminals are equal. As a quick intro to this concept please follow this example:
**broken link removed**
Assume the op-amp has it's power supply infinite positive voltage and infinite negative voltage so that it never saturates. You apply a voltage at Vin. Since the op-amp terminals are ideally infinite impedance, no current flows into the opamp input so no voltage drop is formed across the resistor so the voltage at Vin equals the voltage at the non-inverting input. Got it?

Now the op-amp will try and output enough voltage to make the voltages at the two input terminals equal. If the voltage at the non-inverting input is more positive than the one at the inverting input, the op-amp will SOURCE current (ie. output a positive voltage) and if the inverting is more positive than the non-inverting the op-amp will SINK current (ie. output a negative voltage). It will keep increase (or decreasing) the output voltage with this trend until the input voltages are equal. So let's assume Vin is positive (ie. above ground which is connected to the inverting terminal through resistors). THe op-amp will output a current down through the resistors to ground such that the voltage formed by this current across Rin (and therefore the inverting input voltage) is equal to the non-inverting input voltage.

BUt what's this? THis current also goes through Rf! THat means that the current flowing through the two resistors (remember no current flows into the op-amp inputs) is I=Vin/Rin to cause the voltage at the inverting input to match the voltage at the non-inverting input. BUT this same current is flowing through both resistors and the voltage drop formed across them is Vout = I*(Rin+Rf). ANd since this resistor divider is referenced to ground, then the output voltage is also referenced to ground. If we rearrange and make the two equations equal to each other, we get:

Gain = Vout/Vin = (Rin+Rf)/Rin = 1 + Rf/in

THe gain is positive. WHat is happening is the magnitude of the current is being decided by one resistor, but the output voltage is being decided by the current flowing through a LARGER resistance (the one resistor plus another resistor) so you get a voltage gain. In order for the op-amp to produce a matching voltage at the input terminals (ie. drive enough current to produce the voltage across Rin) it must have a higher voltage at it's output since this extra voltage is being dropped across the other resistor, Rf.

THat is as far as I will go. Please go through this link carefully.

Walk through this:

I appreciate the trouble you went through to explain that, and im sure it makes sense to someone that gets technical electronic jargon, could you please pretend i am ten and know nothing about electronics, it will make it so much easier to understand what you mean, sorry to be an ass but i just dont understand.

 

[Hero999]

I'll try to explain it to you but you might need to do some research on ohm's law before you can understand it.

An op-amp is an amplifier. This means it gets the energy for the output from the power supply. The maximum output voltage is always less than the power supply rails as the output transistors have a certain voltage loss. The output current is also always less than the input current as some current is used by the op-amp.

Voltage gain means the amount of voltage multiplication. If an amplifier has a gain of 2, an input voltage of 2V become 4V. If an amplifier has a gain of -2 and input voltage of 2V becomes -4V. If an amplifier has a gain of 0.5 and input voltage of 2V becomes 1V.

An op-amp is a differential amplifier. This means it amplifies the difference betweeen the + and - inputs. An op-amp has a very high gain, often as high as 1 million.

If there is no negative feedback the op-amp will act as a comparator.

What is a comparator?

A device which outputs a positive votlage (minus a small loss) when the + input is higher than the - input and a negative voltage whent the - input is higher than the + input.

An op-amp does this because it has a very high gain and the limit of the output voltage is the supply voltage.

For example, if the + input is 2.5V and the - input is 2V and the op-amp has a gain of 1 million it will try to push the output to half a million voltage which isn't possible so it sits pretty close to the positive supply voltage.

What is negative feedback?

When a proportion of the output which the opposite sign is fed back to the input.

Now let's look at the inverting amplifier.
**broken link removed**
Notice how the input goes to the - input and the + input is connected to 0V?

There are two resistors between the output an the input.

Lets set them to both 1k.

Remember the amplifier is an inverting amplifier so the output has the opposite sign as the input.

If we put 1V into the input, the amplifier will look at the difference between the + and - inputs and multiply the diffence between the two signals by minus 1 million. Now you'd expect the output to go the the negative supply rail but in practice Rf allows some of the output signal back to the i input which cancels out the input by a factor of 1 million times Rf over Rin (remember the potential divider formula). This means that the 1V input is reduced to just one millionth of a volt and the output sits at -1V.

Take a look at the unity gain buffer.

**broken link removed**
All of the ouput is fed back in to the - input. Suppose we have an input voltge of 1V and the op-amp's open loop gain (this gain without negative feadback) is 1 million. The voltage at the - input will be 1 - 1 millionth of a volt or 0.999999V, for all intents and putposes this is near enough 1V for most applications.

Let's take a look at the non-inverting amplifier.
**broken link removed**
If you reduce the negative feedback by using a potential divider that only feeds a fraction of the output back to the -input the gain is increased.

Hopefully these formulas will make more sense to you now.

I don't have the time at the moment to explain it any further at the moment but if you have any questions I'll try to answer them.

All I can says is make sure you understand the potential divider theory and playing around with simulators can also help.

 

[Roff]

has a good tutorial on electronics. Volume III, Chapter 8 is on op amps, but, judging on your comments, you should start on Vol. I. I'm afraid you will never understand the op amp chapter with the background you currently have.